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[Solved]: af_(j)^(')=(f_(j-2)-4f_(j-1)+3f_(j))/(2 Delta
Home / Expert Answers / Advanced Math / af-j-39-f-j-2-4f-j-1-3f-j-2-delta-delta-2-3-f-j-3-o-delta-3-f-j-39-pa826

(Solved): af_(j)^(')=(f_(j-2)-4f_(j-1)+3f_(j))/(2\Delta )+(\Delta ^(2))/(3)f_(j)^((3))+O(\Delta ^(3)) f_(j)^(' ...



af_(j)^(')=(f_(j-2)-4f_(j-1)+3f_(j))/(2\Delta )+(\Delta ^(2))/(3)f_(j)^((3))+O(\Delta ^(3)) f_(j)^(')=(-3f_(j)+4f_(j+1)-f_(j+2))/(2\Delta )-(\Delta ^(2))/(3)f_(j)^((3))+O(\Delta ^(3)) by algebraic manipulations of Taylor series expansions of function f. One-sided difference stencils for the left and right edges of the computational domain bf(x) at x=x_(j) by finding the coefficients a_(o),a_(1),a_(2) of two parabolas of the form tilde(f)(x)=a_(o)+a_(1)x+a_(2)x^(2) that pass through set of points (x_(j-2),f_(j-2)),(x_(j-1),f_(j-1)),(x_(j),f_(j)) and (x_(j),f_(j)),(x_(j+1),f_(j+1)),(x_(j+2),f_(j+2)), respectively, and evaluating their first derivatives with respect to x at x=x_(j).


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